Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
3. Bisectors in Triangles
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Exercise 22 Page 306

The incenter is the point of concurrency of the angle bisectors.

m∠ DGE = 130^(∘)
m∠ DGF = 120^(∘)
m∠ EGF = 110^(∘)

Practice makes perfect

Let's use the given diagram and information to find the measures of the other three angles. We will add the given measure, m∠ DEF=60, to the diagram to help us visualize the situation a little better.

We are also given the relationship between two of the angles. m∠ EFD =2 m∠ EDF We want to find the measures of ∠ DGE, ∠ DGF, and ∠ EGF. We know that the sum of the angles in a triangle is 180^(∘). We can use this fact to create an equation for the measures of the angles of the large triangle. m∠ DEF + m∠ EFD + m∠ EDF = 180 ^(∘) Let's substitute the values that we know into the equation. 60^(∘) + ( 2 m∠ EDF)+ m∠ EDF = 180^(∘) Now we can solve for the measure of angle EDF.
60 + (2 m∠ EDF)+ m∠ EDF = 180
60 + 3 m∠ EDF = 180
3 m∠ EDF = 120
m∠ EDF = 40
We can now use the given relation to find the measure of ∠ EFD. m∠ EFD =& 2 m∠ EDF m∠ EFD =& 2 ( 40^(∘)) = 80^(∘) Since the problem states that G is the incenter of the triangle, we know that DG, FG, and EG are angle bisectors. Angle bisectors divide the angle into two congruent angles.
If we look at the triangle as three smaller triangles, we can use the interior sums to find the remaining missing angles. 20^(∘) + 30^(∘) + m∠ DGE = 180^(∘) Let's solve for m∠ DGE.
20+30+m∠ DGE = 180
50+m∠ DGE = 180
m∠ DGE = 130
We can use the same process to find the other two angles. 20 + 40 + m∠ DGF = 180 → m∠ DGF = 120^(∘) 30 + 40 + m∠ EGF = 180 → m∠ EGF = 110^(∘) We have now found the measures of all three of the angles.